One of the requirements for the existence of Fourier transform of $f(x)$ is that:
$\int_{-\infty}^{\infty} |f(x)| dx $ exists.
However, the table says that the Fourier transform of constant functions (\emph{i.e.}, $f(x)=1$) do exist and it is $\delta(k)$ although $\int_{-\infty}^{\infty} 1 dx = \infty$ .
Could anyone can help me to understand this? Thanks in advance.
On
I think it might be good to think the dirac as the limit of a some know function. For example when we take a zero mean Gaussian density $f(x;0, \sigma)$ with variance $\sigma^2$, we know that when $\lim_{\sigma\rightarrow 0}f(x;0, \sigma)=\delta(x)$. This is one type of the definition of the dirac delta function.
When we take the fourier transform of this function $$\int_{-\infty}^{\infty}f(x)e^{itx}\mbox{d}x=e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}$$ and when $\sigma\rightarrow\infty$ we have $$|e^{i\mu t}e^{-\frac{1}{2}(\sigma t)^2}|\rightarrow 1$$
Another way would be to take a rectangular function on $[-t,t]$ at the frequency domain $t$ and let $t\rightarrow\infty$. At the time domain $x$ one would get a sinc function with a single main lobe. The number of zero crossings will increase, the power of side lobes will decrease and eventually go to zero, what remains will be a dirac delta function.
Mathematically, if absolute summability is not satisfied we also get something which is strangely defined; something which has an infinite power at frequency $t=0$, from Perseval's energy preservation rule.
The Fourier transform defined by an ordinary (Riemann or) Lebesgue integral only exists when $f \in L^1$.
It is however possible to extend the definition to tempered distributions (for example, every locally integrable function that "doesn't grow too fast" can be identified with a tempered distribution). The Fourier transform of such a thing is not in general a function though, as witnessed by your example.